Commutator of Lorentz boost generators : visual interpretation

I have always struggled to visualize the correctness of the commutation relation for the generators of the boost in the Lorentz group. We have $$[K_i,K_j] = i \epsilon_{ijk} L_k$$ I fail to picture this. For definiteness' sake, let's take a point $\vec{x}$ in my coordinate system, lying in the $O_{xy}$ plane. The difference between boosting first in the $x$ and then in the $y$ direction, and boosting first in the $y$ and then in the $x$ direction ammounts simply to a rotation in the $Oxy$ plane, since $$[K_1,K_2] = iL_3$$ I must be doing something wrong when I draw this on a paper. I realize that this holds at the infinitesimal level, but a simple drawing should be able to realize this principle 'at first order' shouldn't it ? My problem is that when I draw it, I end up with the same situation regardless of which direction I boost into first.

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Thomas precession might be a good example. – jinawee Jun 19 '13 at 21:04
$K_1$ exchanges $x$ and $t$, $K_2$ exchanges $y$ and $t$, $L_3$ is a $\pi/2$ rotation in the $xy$ plane. It is a tri-dimensional problem. For instance, begin with the initial point $x,y,t = 1,0,0$ – Trimok Jun 19 '13 at 21:07
You might find this paper helpful: Rhodes and Semon, Am. J. Phys. 72(7)2004, bates.edu/~msemon/RhodesSemonFinal.pdf . They have a way of representing boosts on a disk. I have a shorter exposition here lightandmatter.com/genrel (subsection 2.5.3) on how this implies that the product of two boosts isn't a pure boost. – Ben Crowell Jun 19 '13 at 23:44
@BenCrowell Thank you, very nice readings. I find your(?) book very good. The 'velocity disk' example in section 2.5.3 is convincing. – Mathusalem Jun 20 '13 at 7:00

1 Answer

Visualizing this is, in my opinion, pretty tricky because boosts generally act in four dimensions (time included!) and it's hard to draw 4D on a piece of paper. Fortunately, for this question, you can get around this artistic limitation.

Here's what I did right after reading your question to attempt to visualize what's going on. I encourage you to try to draw each step. The picture I ended up drawing is at the end.

If we order our coordinates in the standard way $(t,x,y,z)$, then boosts with boost parameter $\beta$ in the $x$ and $y$ directions are given by $$\Lambda_x(\beta) = \left( \begin{array}{cccc} \gamma & -\gamma\beta & 0 & 0 \\ -\gamma\beta & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right), \qquad \Lambda_y(\beta)=\left( \begin{array}{cccc} \gamma & 0 & -\gamma\beta & 0 \\ 0 & 1 & 0 & 0 \\ -\gamma\beta & 0 & \gamma & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right)$$ We can now attempt to apply these transformations to some point in the $x$-$y$ plane in different orders and draw the results. The key to visualizing the math I'm about to do is to note that both of these boosts do nothing to the $z$ coordinate of any spacetime point because there is a $1$ in the lower right hand corner of each transformation matrix with zeros to its left and above. As a result, we can ignore the $z$ axis and draw spacetime diagrams in which the time axis is vertical and the $x$-$y$ plane is horizontal.

For simplicity, let's see what happens to the point $e_x=(0,1,0,0)^t$. Here I will use a superscript $t$ for transpose since I want to emphasize that points in spacetime will be written as column vectors. The point $e_x$ is simply the three-dimensional unit vector along the $x$-axis at time zero. What happens when we apply a small boost in the $y$-direction followed by a small boost in the $x$-direction? Well, let's compute; $$\Lambda_x(\beta)\Lambda_y(\beta)e_x\approx e_x+(-\beta, \beta^2/2, 0, 0)^t$$ The $\approx$ symbol means that I have simply dropped lower order terms in the Taylor expansion in $\beta$ about $\beta = 0$. The point is translated backward in time and forward in $x$. Now what happens when we apply the transformations in the other order and then expand the result? $$\Lambda_y(\beta)\Lambda_x(\beta)e_x\approx e_x+(-\beta, \beta^2/2, \beta^2, 0)^t$$ The same thing happens to the time and $x$ components as in the first order, but now the point is also shifted forward in $y$ a bit. When we subtract these two, we therefore get $$\Big(\Lambda_x(\beta)\Lambda_y(\beta)-\Lambda_y(\beta)\Lambda_x(\beta)\Big)e_x \approx (0,0,-\beta^2,0)^t$$ Does this agree with what a rotation in $z$ would do? Well a rotation in $z$ by an angle $\phi$ looks like this: $$R_z(\phi)=\left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & \cos \phi & \sin \phi & 0 \\ 0 & -\sin \phi & \cos \phi & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right)$$ If you Taylor expand this about $\phi=0$, you'll get $$R_z(\phi) \approx I + \left( \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 0 & \phi & 0 \\ 0 & -\phi & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right)$$ So a rotation in $z$ acts on the point $e_x$ to give $$R_z(\phi) e_x \approx e_x + (0,0,-\phi,0)^t$$ In other words, it subtracts a bit from the $y$-coordinate. But this subtraction is precisely what happened in the case of of the two boosts acting in different orders once we identify $\phi =\beta^2$!

I know that this was all rather mathematical, but I can't see a way of getting around doing some math so as to be able to draw what happens to a point you choose under a given small boost or rotation.

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Thank you very much. I can clearly see it works. The thing is I was associating a boost too much with a simple momentum addition in a direction, and had completely forgotten to factor in how the boost also acts on time. – Mathusalem Jun 19 '13 at 21:23
@Mathusalem Sure thing. – joshphysics Jun 19 '13 at 21:39